Professor in Process Control
Process Dynamic and Control covers
The objectives are to:
In its simplest form, a mathematical model is nothing more than equations that relate the values of variables, it can be decomposed into four basic components: input variables, output variables, parameters, and operators. Output variables are the unknown quantities that the model is designed to deduce from values of the known inputs. The operators in the model define the mathematical manipulations required to compute the value of the output from the inputs and parameters.
The dynamic is ( Of or relating to energy or to objects in motion, Merriem Webster Dictionary) the time varying behavior of processes. Most of chemical process behaviors are dynamically continuous changes respecting to inputs. The process in steady state change indicates where the process is going and the dynamic characteristic of the process indicates what dynamic path it will take. The dynamic of the process is usually represented by difference, differential, parametric, or state space models. In the following, these dynamical models are discussed.
A dynamic mathematical model can be represented in difference and differential equations. Usually, the model developed based on first principle is differential equations, they can be lumped or distributed equations. A model developed from a given input-output data is difference equations, which are usually represented in parametric model.
Suppose there is defined a sequence of points, perhaps representing discrete equally space time point, indexed by \(k\). Suppose also that there is a value \(y(k)\) (a real number) associated with each of these point. A difference equation is an equation relating the value \(y(k)\), at point \(k\), to value at other (usually neighboring) points. A simple example is \[y(k+1)=ay(k) \qquad k=0,1,2,\dots \] Difference equation may, however, be much more complicated than this. A solution of a difference equation is a function \(y(k)\) that reduces the equation to an identity. For example, corresponding to the first-order equation above, the function \(y(k)=a^k\) reduces the equation to an identity, since \(y(k+1)=a^{k+1}=aa^k=ay(k)\).
Usually, the first order difference equation is represented to take into account the bias of the function, which is known as the error compensate of the function. The equation is \[y(k+1)=ay(k)+b\] linear, has a constant \(a\), and a constant forcing term \(b\). The solution of the simple representation of the identity equation is by finding the simplest relation and correlation of the value in the neighborhood. From the equation above, the value of \(y\) at an initial point \(k_0\), say \(k_0=0\) and specify \(y(0)=y_0\). The following corresponding values are depend on the initial value, which in sequence is represented as \[ \begin{align} y(k_0=0)=&y_0\\ y(k_1=k_0+1=1)=&ay(0)+b=ay_0+b\\ y(k_2=k_1+1=1+1=2)=&a(y_1)+b=a(ay_0+b)+b=a^2y_0+ab+b\\ \vdots\\ y(k_n)=&a^ny_0+(a^{n-1}+a^{n-2}+\dots+a+1)b \end{align} \] The first term of the right hand side of the equation depends on the initial value, \(y_0\). However, the second term corrects the sequence of errors. For \(a=1\), the expression reduces simply to \[y(k_n)=y_0+nb.\] However, in the real facing problem, the weighted parameter is not necessarily equal to 1. Therefore, if \(a\neq1\), the expression can be somewhat simplified by collapsing the geometric series using \[ 1+a+a^2+\dots+a^{n-1}=\frac{1-a^n}{1-a} \] Therefore, the desired solution in closed-form is \[ y(k_n)=\left\{ \begin{array}{ll} y_0+nb, & a=1 \\ a^ny_0+\frac{1-a^n}{1-a}b, & a\neq1 \end{array} \right. \]
Suppose there is an interval, \(t_0\leq t\leq t_1\), representing an interval of continuous time. Suppose also that there is a value \(y(t)\) associated with each point \(t\) in that interval. Then \(y(t)\) is a function defined on the interval. A differential equation is an equation connecting such a function and some of its derivatives. A simple example is the equation \[\frac{dy}{dt}=ay,\quad y(0)=y_0\] which is an autonomous system. To drive the system is by introducing a force variable \(b\) such as \[\frac{dy}{dt}=ay+b,\quad y(0)=y_0\] The solution for the autonomos system is \[y(t)=y_0e^{at} \] and solution for non-autonomous system is given \[ y(t)=-\frac{b}{a}+\left(y_0+\frac{b}{a}\right)e^{at} \]
In general the system can be modeled by arranging that the system is influenced by a receiving and releasing material or energy and/or system's characteristics. This can be a chemical reaction or something else. \[\begin{bmatrix} \text{Rate of} \\ \text{Accumulation} \end{bmatrix}=\begin{bmatrix} \text{Rate of Entering} \\ \text{the systems} \end{bmatrix}-\begin{bmatrix} \text{Rate of Leaving} \\ \text{the systems} \end{bmatrix}+\begin{bmatrix} \text{Rate of Generation } \\ \text{or Consumption within system} \end{bmatrix}\]
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\[ u(t)=k_c\left(e(t)+\frac{1}{t_i}\int_0^te(\tau)d\tau+t_d\frac{de(t)}{dt}\right) \] where
Take a first-order plus dead time (FOPDT) plant \[G_p(s)=\frac{k_p}{\tau_ps+1}e^{-ls}\] apply a unit step \(\Delta u=1\) and \(t=0\). The time-domain output (for \(t\ge l\)) is \[y(t)=k_p\left(1-e^{-\frac{t-l}{\tau_p}}\right),\quad t\ge l\] and \(y(t)=0 \) for \( t < l \). Let for example the plant has \(k_p=2, \tau_p=5 \)s and \( l=1 \)s, therefor \(t\ge 1\) is= \[y(t)=2\left(1-e^{-\frac{t-1}{5}}\right) \] The classical reaction-curve (open-loop) method uses the tangent at the beginning of the response. For the FOPDT model above: